I have yet to see any problem, however complicated, which, when looked at in the right way did not become still more complicated, Paul Anderson.
Ned, I would love to stand here and talk with you—but I’m not going to, Phil Connors (Bill Murray), Groundhog Day
I don’t have to take this abuse from you; I’ve got hundreds of people dying to abuse me, Dr. Peter Venkman (Bill Murray), Ghostbusters
Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$
It involves (e.g., $\frac{dy}{dx} = 3x +5y$):
Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:
These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.
If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.
The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:
Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}), meaning that it satisfies the initial condition y(x_{0}) = y_{0}.
This theorem ensures that under these conditions, the solution exists and is unique near x = x_{0}.
Definition. A first order autonomous ordinary differential equation (ODE) is a differential equation in which the rate of change of a dependent variable depends solely on y itself, and not explicitly on the independent variable, typically denoted by t. In mathematical terms, it has the general form: $\frac{dy}{dt}=f(y)$.
Here, f(y) is a function that depends only on y, and there is no explicit dependence on the independent variable t. This absence of explicit dependence on t is what characterizes the equation as autonomous.
Because the right-hand side of the equation depends only on y, autonomous ODEs exhibit unique properties that distinguish them from non-autonomous equations:
While autonomous ODEs are separable, the integration may not always be straightforward due to:
In such situations, qualitative analysis becomes a valuable tool to gain insights into the behavior of the solutions without solving the equation analytically.
In qualitative analysis, we study the direction field of the ODE. This field helps visualize the behavior of solutions without explicitly solving the equation.
A direction field (or slope field) is a graphical representation of the ODE that shows the slope $\frac{dy}{dt}$ at various points in the (t, y) plane. By plotting small line segments with appropriate slopes at several points, we can visualize how solutions to the ODE behave.
Behavior Based on f(y). When f(y) > 0, the solution (the slope $\frac{dy}{dt}$ is positive) y(t) is increasing at that point. When f(y) < 0, the solution (the slope $\frac{dy}{dt}$ is negative) y(t) is decreasing at that point.
A critical point or equilibrium solution occurs where f(y) = 0. At these points, the rate of change $\frac{dy}{dt} = 0$,meaning that the solution remains constant over time.
If y_{0} is a critical point, then y(t) = y_{0} is a solution to the differential equation because: $\frac{dy}{dt}|_{y=y_0}=\frac{dy_0}{dt}=f(y_0)=0$. The left hand size is zero because the derivative of a constant is zero, and the right hand side is zero because f(y_{0}) = 0 by definition of a critical point.
To analyze an autonomous ODE, follow these steps:
Find the critical points. Solve f(y) = 0 to identify where the solutions remain constant (i.e., equilibrium solutions).
Analyze the sign of f(y) and draw the Phase Line. Sketch or plot f(y) versus y to visualize where f(y) is positive, negative, or zero.
When f(y) > 0, the slope $\frac{dy}{dt}$ is positive, meaning that the solution y(t) is increasing. When f(y) < 0, the slope $\frac{dy}{dt}$ is negative, meaning that the solution y(t) is decreasing.
Determine Stability of Critical Points: Stable Equilibrium (Solutions near the critical point move towards it as t increases. The point is attracting), Unstable Equilibrium (Solutions near the critical point move away from it as t increases. The point is repelling), or Semi-Stable Equilibrium (The critical point is stable on one side and unstable on the other).
Interpret the Results. Discuss the long-term behavior of solutions. Determine whether solutions approach a critical point, diverge to infinity, or exhibit other behaviors based on initial conditions.
Consider initial conditions. Analyze how different initial values y(0) affect the solution’s trajectory and its tendency to move towards or away from critical points.
Autonomous Nature. This equation is a first order linear autonomous ODE because it relates the derivate $\frac{dy}{dt}$ to the function y(t) itself and does not explicitly depend on the independent variable t.
Finding Critical Points. Equilibrium points occurs where $\frac{dy}{dt} = 0$, $\frac{dy}{dt} = 0 ↭ 1-y^2 = 0↭ (1-y)(1+y) = 0 ↭ y = ±1$.
Analyzing the Sign of f(y). To understand this differential equation, we analyze the sign of $\frac{dy}{dt}$ for values of y above and below the critical point.
Drawing the Phase Line. Arrows indicating the direction of y(t) point upwards (towards y = 1, solutions increase) from -1 < y < 1, indicating attraction, so y = 1 is a stable equilibrium point.
Slopes Between y = −1 and y = 1: (1) Start at Zero at y = −1 ($\frac{dy}{dt} = 0$). (2) Between y = -1 and y = 0, slopes increase from 0 to a maximum of 1 ($\frac{dy}{dt}=1-y²\bigg|_{y=0}=1-0 = 1$) at y = 0. (3) Between y = 0 and y = 1, slopes decrease back from its maximum 1 to zero (y = 1). Solutions increase fastest near y = 0. The rate of increase slows down as y approaches y = 1.
Below y = -1 (y < -1) and above y = 1 (y > 1), slopes are negative, solutions y(t) decreases as t increases, arrows point downwards, that is, away from y = -1 and towards y = 1, respectively, hence y = -1 is a unstable equilibrium point.
Plots of y(t) for different initial conditions illustrate the convergence towards y = 1 for ∣y_{0}| < 1 and y_{0} > 1, and divergence for y_{0} < -1 (Refer to Figure i for a visual representation and aid in understanding it)
This is an autonomous, separable first-order ordinary differential equation (ODE): $\frac{dy}{1-y^2} = dt$.
We perform partial fraction decomposition: $\frac{dy}{1-y^2} = \frac{1}{2}(\frac{1}{1+y}+\frac{1}{1-y})$
$\frac{1}{1.y^2} = \frac{1}{(1-y)(1+y)} = \frac{A}{1-y} + \frac{B}{\frac{1}{1+y}}↭ 1 = A(1+y)+B(1-y) = A(1)+A(y)+B(1)−B(y)=(A+B)+(Ay−By)↭[\text{Group like terms}] A + B = 1, A - B = 0$. From (ii), A = B. From (1), 2A = 1, $A = \frac{1}{2}, B = A = \frac{1}{2}$.
Integrating both sides: $\int \frac{1}{2}(\frac{1}{1+y}+\frac{1}{1-y})dy = \int dt$.
This yields: $\frac{1}{2}ln|1+y| -\frac{1}{2}ln|1-y| = t + C ↭ ln|1+y| -ln|1-y| = 2(t + C) ⇒[\text{Combining logarithms:}] ln(\frac{1+y}{1-y}) = 2(t+C) ⇒[\text{Exponentiating both sides of the equation}] \frac{1+y}{1-y} = e^{2(t+C)} = Ae^{2t}$ where A = e^{2C}.
$\frac{1+y}{1-y} = Ae^{2t} ↭ 1+y = Ae^{2t}(1-y)↭ 1 + y = Ae^{2t} -Ae^{2t}y ⇒[\text{Collect like terms}] y(1+Ae^{2t}) = Ae^{2t} -1⇒[\text{Solve for y}] y = \frac{Ae^{2t} -1}{Ae^{2t} +1}$. This is the explicit solution for y(t).
The constant A can be found using an initial condition y(0) = y_{0}, $y_0 = \frac{A-1}{A+1}↭ (A+1)y_0 = A -1 ↭ Ay_0 + y_0 = A - 1 ↭ Ay_0 -A = -y_0 -1 ↭ A(y_0 -1) = -y_0 -1 ↭ A = \frac{-y_0-1}{y_0-1}↭ A = \frac{y_0+1}{1-y_0}$
The solution to the differential equation is: $\frac{Ae^{2t} -1}{Ae^{2t} +1}$ where $A = \frac{y_0+1}{1-y_0}$
Analyzing the Limits, $\lim_{t \to ∞} y(t) = \lim_{t \to ∞} \frac{Ae^{2t} -1}{Ae^{2t} +1} = 1$ (the terms -1 and 1 become negligible compared to Ae^{2t}).
Regardless of the value of A (provided A > 0), y(t) approaches 1 as t → ∞, confirming that y = 1 is a stable equilibrium. Besides, $\lim_{t \to -∞} y(t) = \lim_{t \to -∞} \frac{Ae^{2t} -1}{Ae^{2t} +1} = \frac{0-1}{0+1} = -1$, which aligns with the instability of y = −1 in forward time (y = -1 is an unstable equilibrium) but shows attraction in backward time.
Let’s explore how to model and analyze the behavior of a bank account that is simultaneously accruing interest and experiencing consistent withdrawals due to embezzlement. This scenario helps illustrate how differential equations can be used to understand real-world financial situations.
Let’s consider the following scenario where:
Suppose a bank teller is illicitly withdrawing funds from a customer’s bank account without their knowledge or voluntary consent. The customer continues to accrue interest on their account balance at a continuous rate r, but the ongoing embezzlement at rate w affects the overall growth of the account. We aim to model the evolution of the account balance over time and understand under what conditions the balance will increase or decrease.
The rate of change of the account balance y(t) over time can be expressed as the difference between the interest earned and the amount being embezzled: $\frac{dy}{dt} = ry -w$ where:
Analysis of the ODE
Autonomous Nature: This equation is a first-order linear autonomous ODE because it relates the derivative $\frac{dy}{dt}$ to the function y(t) itself and does not explicitly depend on the independent variable t.
An autonomous ODE is one where the rate of change of the dependent variable depends solely on the dependent variable itself and not explicitly on the independent variable.
Finding Critical Points (Equilibrium Solutions): Critical points occur when the rate of change of the dependent variable is zero: $\frac{dy}{dt} = 0 ⇒ ry -w = 0 ↭[\text{Solving for y:}] y = \frac{w}{r}$ (Refer to Figure 2 for a visual representation and aid in understanding it). Hence, $y = \frac{w}{r}$ is the equilibrium solution or critical point.
At this balance, the amount of interest earned exactly offsets the amount being embezzled, resulting in a net zero change in the account balance over time.
Arrows indicating the direction of y(t). Upwards for y > w/r, and downwards for y < w/r
Behaviour of the solution. To understand how the account balance changes over time, we analyze the sign of $\frac{dy}{dt}$ for values of y above and below the critical point.
If the bank account has more than w/r (the critical point), the account balance increases over time because the interest earned per unit time exceeds the embezzlement rate. The customer may not notice the embezzlement immediately due to the increasing balance.
If the bank account has less than w/r (the critical point), the account balance diminishes because the losses due to embezzlement outweigh the interest gains. The account may eventually deplete, potentially leading to financial issues.
Separate Variables: $\frac{dy}{dt} = ry-w ↭ \frac{dy}{ry-w} = dt$
Integrate Both Sides: $\int \frac{dy}{ry-w} = \int dt ↭[u = ry -w⇒ du = rdy] \frac{1}{r}\int \frac{du}{u} = t + C ↭ \frac{1}{r}ln|u| = t + C ↭ \frac{1}{r}ln|ry-w| = t + C$
Solve for y. Multiply both sides by r: ln|ry -w| = rt + rC. Exponentiate both sides to eliminate the natural logarithm: $|ry -w| = e^{rt}e^{rC}↭ ry -w = ±e^{rC}e^{rt} ↭ ry -w = ±Ae^{rt}$ where $A = e^{rC}$ is a constant ↭ $ry = w ±Ae^{rt} ↭ y = \frac{w}{r} ±\frac{A}{r}e^{rt} ↭ y = \frac{w}{r} + Ce^{rt}$ where $C = ±\frac{A}{r}$
Apply initial conditions, y(0) = y_{0}, $y_0 = \frac{w}{r}+ Ce^{r·0}↭ y_0 = \frac{w}{r}+ C ⇒ C = y_0 -\frac{w}{r}$
Final Explicit Solution: $y(t) = \frac{w}{r}+(y_0-\frac{w}{r})e^{rt}$
Interpretation: